Linear-Algebra-Done-Right · Unit 28 · Video 3 · Interactive Practice

Top-Degree Alternating Forms: A One-Dimensional Space

IKey Formulas

Formula	Name	What it says
$\alpha(v_1, \dots, v_n) = \alpha(e_1, \dots, e_n) \displaystyle\sum_{(j_1, \dots, j_n) \in \operatorname{perm} n} \operatorname{sign}(j_1, \dots, j_n)\, b_{j_1, 1} \cdots b_{j_n, n}$	The formula ( $n = \dim V$ , $v_k = \sum_j b_{j,k} e_j$ )	One number, $\alpha(e_1, \dots, e_n)$ , fixes $\alpha$
$\dim V_{\text{alt}}^{(n)} = 1$	Top degree	Every such form is a multiple of one nonzero form
$\alpha(v_1, \dots, v_n) = \displaystyle\sum_{\text{perm } n} \operatorname{sign}(j_1, \dots, j_n)\, \varphi_{j_1}(v_1) \cdots \varphi_{j_n}(v_n)$	A nonzero form from the dual basis	Alternating, and $\alpha(e_1, \dots, e_n) = 1$
$\alpha \ne 0 \Rightarrow \big(\alpha(e_1, \dots, e_n) \ne 0 \iff e_1, \dots, e_n$ independent $\big)$	The independence test	Lists of length $n = \dim V$

Key Insight: of the $n^n$ terms from multilinearity, only the $n!$ with all indices different survive, and sorting each into $e_1, \dots, e_n$ costs exactly its sign.

IITwenty-Seven Terms, Six Survivors

For $n = 3$ , multilinearity gives $27$ terms; repeated indices kill $21$ , and sorting the other $6$ costs their signs.

Vectors Step 1

IIIThe Case $n = 2$ and the Independence Test

On a $2$ -dimensional space, $\alpha(v_1, v_2) = \alpha(e_1, e_2)\,(b_{1,1} b_{2,2} - b_{2,1} b_{1,2})$ .

α(e₁, e₂) 1

IVQuiz Questions

Problem 1 · The Formula for $n = 2$

Given: $\dim V = 2$ , a basis $e_1, e_2$ , and $\alpha \in V_{\text{alt}}^{(2)}$ with $\alpha(e_1, e_2) = 3$ . Find $\alpha(2e_1 + e_2,\ e_1 + 4e_2)$ .

✅ Correct!

3\,(2 \cdot 4 - 1 \cdot 1) = 21

❌ Not quite.

2 \cdot 4 - 1 \cdot 1 = 7

is the sum; it still has to be multiplied by

\alpha(e_1, e_2) = 3

❌ Not quite.

\alpha(e_2, e_1) = -\alpha(e_1, e_2)

, so the cross term enters with a minus sign:

b_{1,1}b_{2,2} - b_{2,1}b_{1,2}

❌ Not quite. The identity permutation

(1, 2)

has sign

+1

; it is the term

b_{2,1} b_{1,2}

that is subtracted.

❌ Not quite. Expand both slots and drop the terms with a repeated basis vector.

Show solution

Here $b_{1,1} = 2$ , $b_{2,1} = 1$ (coordinates of $v_1$ ) and $b_{1,2} = 1$ , $b_{2,2} = 4$ (coordinates of $v_2$ ). Expanding,

\alpha(v_1, v_2) = b_{1,1}b_{2,2}\,\alpha(e_1, e_2) + b_{2,1}b_{1,2}\,\alpha(e_2, e_1) = (2 \cdot 4 - 1 \cdot 1)\,\alpha(e_1, e_2),

since $\alpha(e_1, e_1) = \alpha(e_2, e_2) = 0$ and $\alpha(e_2, e_1) = -\alpha(e_1, e_2)$ . So $\alpha(v_1, v_2) = 7 \cdot 3 = 21$ .

Problem 2 · Counting Survivors

Given: $\dim V = 4$ and $\alpha \in V_{\text{alt}}^{(4)}$ . Expanding $\alpha(v_1, v_2, v_3, v_4)$ by linearity in all four slots gives terms $b_{j_1,1} b_{j_2,2} b_{j_3,3} b_{j_4,4}\,\alpha(e_{j_1}, e_{j_2}, e_{j_3}, e_{j_4})$ . How many terms are there, and how many can be nonzero?

✅ Correct!

4^4 = 256

index lists, and only the

4! = 24

with all indices different avoid a repeated basis vector.

❌ Not quite. Each of the

4

slots contributes its own sum over

4

indices, so there are

4 \cdot 4 \cdot 4 \cdot 4

terms.

❌ Not quite.

232 = 256 - 24

is the number of terms with a repeated index, which are the ones that vanish.

❌ Not quite. Signs do not make terms vanish; every permutation term can be nonzero.

Show solution

Linearity in slot $k$ turns $v_k = \sum_{j} b_{j,k} e_j$ into a sum over $j_k \in \{1, 2, 3, 4\}$ , so there are $4^4 = 256$ terms, one for each index list $(j_1, j_2, j_3, j_4)$ .

If two indices are equal, the same basis vector sits in two slots and the term is $0$ because $\alpha$ is alternating. The index lists with all entries different are the permutations of $1, 2, 3, 4$ : there are $4! = 24$ of them.

Problem 3 · Two Forms on One Line

Given: $\dim V = 3$ , a basis $e_1, e_2, e_3$ , and $\alpha, \alpha' \in V_{\text{alt}}^{(3)}$ with $\alpha(e_1, e_2, e_3) = 2$ and $\alpha'(e_1, e_2, e_3) = -6$ . For some $v_1, v_2, v_3$ , $\alpha(v_1, v_2, v_3) = 5$ . Find $\alpha'(v_1, v_2, v_3)$ .

✅ Correct!

\alpha' = -3\alpha

because they agree up to that factor on a basis, so

\alpha'(v_1, v_2, v_3) = -3 \cdot 5 = -15

❌ Not quite. The ratio is

\alpha'/\alpha = -6/2 = -3

, not

2/(-6)

❌ Not quite. The two forms differ by a factor, not by an added constant.

❌ Not quite. Keep the sign of

\alpha'(e_1, e_2, e_3) = -6

: the factor is

-3

❌ Not quite. Apply the formula to both forms with the same basis.

Show solution

By the formula, for any $v_1, v_2, v_3$ ,

\alpha'(v_1, v_2, v_3) = \alpha'(e_1, e_2, e_3)\,S, \qquad \alpha(v_1, v_2, v_3) = \alpha(e_1, e_2, e_3)\,S,

with the same sum $S$ (it depends only on the vectors and the basis). So $\alpha' = \frac{-6}{2}\alpha = -3\alpha$ , and $\alpha'(v_1, v_2, v_3) = -3 \cdot 5 = -15$ .

Problem 4 · When Is the Value Zero?

Given: a nonzero $\alpha \in (\mathbf{R}^2)_{\text{alt}}^{(2)}$ and a real number $t$ . For which $t$ is $\alpha\big((1, t), (t, 4)\big) = 0$ ?

✅ Correct! The value is

\alpha(e_1, e_2)(4 - t^2)

with

\alpha(e_1, e_2) \ne 0

, so it vanishes exactly when

t^2 = 4

❌ Not quite.

t = -2

also works:

(1, -2)

and

(-2, 4)

are dependent, since

(-2, 4) = -2(1, -2)

❌ Not quite. The sum is

1 \cdot 4 - t \cdot t = 4 - t^2

, which vanishes at

t = \pm 2

❌ Not quite. A nonzero top-degree form vanishes exactly on the dependent lists, whichever nonzero form it is.

❌ Not quite. Use the

n = 2

formula with the standard basis.

Show solution

With the standard basis $e_1, e_2$ of $\mathbf{R}^2$ , $v_1 = (1, t)$ has $b_{1,1} = 1$ , $b_{2,1} = t$ and $v_2 = (t, 4)$ has $b_{1,2} = t$ , $b_{2,2} = 4$ . So

\alpha(v_1, v_2) = \alpha(e_1, e_2)\,(1 \cdot 4 - t \cdot t) = \alpha(e_1, e_2)\,(4 - t^2).

Since $\alpha \ne 0$ and $e_1, e_2$ is independent, the test gives $\alpha(e_1, e_2) \ne 0$ . So the value is $0$ exactly when $t^2 = 4$ , that is $t = 2$ or $t = -2$ : the values of $t$ that make the list dependent.

Solved: 0 / 4

Top-Degree Alternating Forms: A One-Dimensional Space

IKey Formulas

IITwenty-Seven Terms, Six Survivors

IIIThe Case n=2n = 2n=2 and the Independence Test

IVQuiz Questions

IIIThe Case $n = 2$ and the Independence Test